?

Average Accuracy: 53.6% → 80.9%
Time: 18.1s
Precision: binary64
Cost: 20164

?

\[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
\[\begin{array}{l} \mathbf{if}\;C \leq 2.1 \cdot 10^{+165}:\\ \;\;\;\;\frac{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{0.005555555555555556}}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{B}{C}, \frac{A \cdot 0}{B}\right)\right)}{\pi}\\ \end{array} \]
(FPCore (A B C)
 :precision binary64
 (*
  180.0
  (/
   (atan (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))
   PI)))
(FPCore (A B C)
 :precision binary64
 (if (<= C 2.1e+165)
   (/ (/ (atan (/ (- (- C A) (hypot B (- A C))) B)) 0.005555555555555556) PI)
   (/ (* 180.0 (atan (fma -0.5 (/ B C) (/ (* A 0.0) B)))) PI)))
double code(double A, double B, double C) {
	return 180.0 * (atan(((1.0 / B) * ((C - A) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / ((double) M_PI));
}
double code(double A, double B, double C) {
	double tmp;
	if (C <= 2.1e+165) {
		tmp = (atan((((C - A) - hypot(B, (A - C))) / B)) / 0.005555555555555556) / ((double) M_PI);
	} else {
		tmp = (180.0 * atan(fma(-0.5, (B / C), ((A * 0.0) / B)))) / ((double) M_PI);
	}
	return tmp;
}
function code(A, B, C)
	return Float64(180.0 * Float64(atan(Float64(Float64(1.0 / B) * Float64(Float64(C - A) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0)))))) / pi))
end
function code(A, B, C)
	tmp = 0.0
	if (C <= 2.1e+165)
		tmp = Float64(Float64(atan(Float64(Float64(Float64(C - A) - hypot(B, Float64(A - C))) / B)) / 0.005555555555555556) / pi);
	else
		tmp = Float64(Float64(180.0 * atan(fma(-0.5, Float64(B / C), Float64(Float64(A * 0.0) / B)))) / pi);
	end
	return tmp
end
code[A_, B_, C_] := N[(180.0 * N[(N[ArcTan[N[(N[(1.0 / B), $MachinePrecision] * N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]
code[A_, B_, C_] := If[LessEqual[C, 2.1e+165], N[(N[(N[ArcTan[N[(N[(N[(C - A), $MachinePrecision] - N[Sqrt[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision] / 0.005555555555555556), $MachinePrecision] / Pi), $MachinePrecision], N[(N[(180.0 * N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision] + N[(N[(A * 0.0), $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / Pi), $MachinePrecision]]
180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}
\begin{array}{l}
\mathbf{if}\;C \leq 2.1 \cdot 10^{+165}:\\
\;\;\;\;\frac{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{0.005555555555555556}}{\pi}\\

\mathbf{else}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{B}{C}, \frac{A \cdot 0}{B}\right)\right)}{\pi}\\


\end{array}

Error?

Derivation?

  1. Split input into 2 regimes
  2. if C < 2.1000000000000001e165

    1. Initial program 59.3%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Simplified75.5%

      \[\leadsto \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)} \]
      Proof

      [Start]59.3

      \[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]

      associate-*r/ [=>]59.3

      \[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}} \]

      associate-*l/ [<=]59.3

      \[ \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)} \]

      associate-*l/ [=>]59.3

      \[ \frac{180}{\pi} \cdot \tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}{B}\right)} \]
    3. Applied egg-rr80.8%

      \[\leadsto \color{blue}{\frac{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{0.005555555555555556}}{\pi}} \]
      Proof

      [Start]75.5

      \[ \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right) \]

      associate-*l/ [=>]75.5

      \[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\pi}} \]

      *-un-lft-identity [=>]75.5

      \[ \frac{180 \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\color{blue}{1 \cdot \pi}} \]

      associate-/r* [=>]75.5

      \[ \color{blue}{\frac{\frac{180 \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{1}}{\pi}} \]

      *-commutative [=>]75.5

      \[ \frac{\frac{\color{blue}{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right) \cdot 180}}{1}}{\pi} \]

      associate-/l* [=>]75.5

      \[ \frac{\color{blue}{\frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\frac{1}{180}}}}{\pi} \]

      associate--r+ [=>]80.8

      \[ \frac{\frac{\tan^{-1} \left(\frac{\color{blue}{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}}{B}\right)}{\frac{1}{180}}}{\pi} \]

      metadata-eval [=>]80.8

      \[ \frac{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\color{blue}{0.005555555555555556}}}{\pi} \]

    if 2.1000000000000001e165 < C

    1. Initial program 10.8%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Simplified10.8%

      \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + B \cdot B}\right)\right)}{\pi}} \]
      Proof

      [Start]10.8

      \[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]

      associate-*r/ [=>]10.8

      \[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}} \]

      sub-neg [=>]10.8

      \[ \frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(\left(C - A\right) + \left(-\sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}\right)}{\pi} \]

      sub-neg [<=]10.8

      \[ \frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}\right)}{\pi} \]

      unpow2 [=>]10.8

      \[ \frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{B \cdot B}}\right)\right)}{\pi} \]
    3. Taylor expanded in C around inf 42.4%

      \[\leadsto \frac{180 \cdot \tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C \cdot B} + -1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
    4. Simplified42.4%

      \[\leadsto \frac{180 \cdot \tan^{-1} \color{blue}{\left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \frac{A \cdot 0}{B}\right)\right)}}{\pi} \]
      Proof

      [Start]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(-0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C \cdot B} + -1 \cdot \frac{A + -1 \cdot A}{B}\right)}{\pi} \]

      fma-def [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \color{blue}{\left(\mathsf{fma}\left(-0.5, \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C \cdot B}, -1 \cdot \frac{A + -1 \cdot A}{B}\right)\right)}}{\pi} \]

      +-commutative [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{\color{blue}{\left({A}^{2} + {B}^{2}\right)} - {\left(-1 \cdot A\right)}^{2}}{C \cdot B}, -1 \cdot \frac{A + -1 \cdot A}{B}\right)\right)}{\pi} \]

      associate--l+ [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{\color{blue}{{A}^{2} + \left({B}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}}{C \cdot B}, -1 \cdot \frac{A + -1 \cdot A}{B}\right)\right)}{\pi} \]

      unpow2 [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{\color{blue}{A \cdot A} + \left({B}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}{C \cdot B}, -1 \cdot \frac{A + -1 \cdot A}{B}\right)\right)}{\pi} \]

      unpow2 [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(\color{blue}{B \cdot B} - {\left(-1 \cdot A\right)}^{2}\right)}{C \cdot B}, -1 \cdot \frac{A + -1 \cdot A}{B}\right)\right)}{\pi} \]

      mul-1-neg [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\color{blue}{\left(-A\right)}}^{2}\right)}{C \cdot B}, -1 \cdot \frac{A + -1 \cdot A}{B}\right)\right)}{\pi} \]

      *-commutative [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{\color{blue}{B \cdot C}}, -1 \cdot \frac{A + -1 \cdot A}{B}\right)\right)}{\pi} \]

      associate-*r/ [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \color{blue}{\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}}\right)\right)}{\pi} \]

      distribute-rgt1-in [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)\right)}{\pi} \]

      associate-*r* [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \frac{\color{blue}{\left(-1 \cdot \left(-1 + 1\right)\right) \cdot A}}{B}\right)\right)}{\pi} \]

      metadata-eval [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \frac{\left(-1 \cdot \color{blue}{0}\right) \cdot A}{B}\right)\right)}{\pi} \]

      metadata-eval [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \frac{\color{blue}{0} \cdot A}{B}\right)\right)}{\pi} \]

      metadata-eval [<=]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \frac{\color{blue}{\left(-1 + 1\right)} \cdot A}{B}\right)\right)}{\pi} \]

      *-commutative [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \frac{\color{blue}{A \cdot \left(-1 + 1\right)}}{B}\right)\right)}{\pi} \]

      metadata-eval [=>]42.4

      \[ \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{A \cdot A + \left(B \cdot B - {\left(-A\right)}^{2}\right)}{B \cdot C}, \frac{A \cdot \color{blue}{0}}{B}\right)\right)}{\pi} \]
    5. Taylor expanded in A around 0 81.9%

      \[\leadsto \frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \color{blue}{\frac{B}{C}}, \frac{A \cdot 0}{B}\right)\right)}{\pi} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification80.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 2.1 \cdot 10^{+165}:\\ \;\;\;\;\frac{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{0.005555555555555556}}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{B}{C}, \frac{A \cdot 0}{B}\right)\right)}{\pi}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy59.7%
Cost20496
\[\begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C + B\right) - A}{B}\right)}{\pi}\\ \mathbf{if}\;C \leq -3 \cdot 10^{-219}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq -1.15 \cdot 10^{-253}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \mathbf{elif}\;C \leq 4 \cdot 10^{-170}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 5.5 \cdot 10^{-93}:\\ \;\;\;\;\frac{\frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{0.005555555555555556}}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{B}{C}, \frac{A \cdot 0}{B}\right)\right)}{\pi}\\ \end{array} \]
Alternative 2
Accuracy80.9%
Cost20164
\[\begin{array}{l} \mathbf{if}\;C \leq 1.15 \cdot 10^{+163}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\mathsf{fma}\left(-0.5, \frac{B}{C}, \frac{A \cdot 0}{B}\right)\right)}{\pi}\\ \end{array} \]
Alternative 3
Accuracy60.3%
Cost13832
\[\begin{array}{l} \mathbf{if}\;B \leq 8.2 \cdot 10^{-292}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C + B\right) - A}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 2.2 \cdot 10^{-209}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{\frac{0.5 \cdot \left(B \cdot B\right)}{A}}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \end{array} \]
Alternative 4
Accuracy45.5%
Cost13576
\[\begin{array}{l} \mathbf{if}\;B \leq -1.75 \cdot 10^{-65}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 7.4 \cdot 10^{-174}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 5
Accuracy50.2%
Cost13576
\[\begin{array}{l} \mathbf{if}\;B \leq 4.8 \cdot 10^{-293}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C + B}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 6.2 \cdot 10^{-174}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 6
Accuracy50.2%
Cost13576
\[\begin{array}{l} \mathbf{if}\;B \leq 1.05 \cdot 10^{-293}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C + B}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 7 \cdot 10^{-174}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 7
Accuracy55.1%
Cost13576
\[\begin{array}{l} \mathbf{if}\;B \leq 4.8 \cdot 10^{-293}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C + B}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 9.5 \cdot 10^{-187}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \end{array} \]
Alternative 8
Accuracy55.1%
Cost13576
\[\begin{array}{l} \mathbf{if}\;B \leq 5.4 \cdot 10^{-292}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C + B}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 8.2 \cdot 10^{-187}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \end{array} \]
Alternative 9
Accuracy55.2%
Cost13576
\[\begin{array}{l} \mathbf{if}\;B \leq 8.2 \cdot 10^{-292}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C + B}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 2.9 \cdot 10^{-183}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{0.5}{\frac{A}{B}}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \end{array} \]
Alternative 10
Accuracy55.1%
Cost13576
\[\begin{array}{l} \mathbf{if}\;B \leq 4.5 \cdot 10^{-296}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C + B}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 10^{-209}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{A \cdot 0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \end{array} \]
Alternative 11
Accuracy60.3%
Cost13576
\[\begin{array}{l} \mathbf{if}\;B \leq 8.5 \cdot 10^{-292}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C + B\right) - A}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 7 \cdot 10^{-208}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{A \cdot 0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \end{array} \]
Alternative 12
Accuracy39.7%
Cost13188
\[\begin{array}{l} \mathbf{if}\;B \leq -2 \cdot 10^{-311}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 13
Accuracy20.8%
Cost13056
\[\frac{180 \cdot \tan^{-1} -1}{\pi} \]

Error

Reproduce?

herbie shell --seed 2023131 
(FPCore (A B C)
  :name "ABCF->ab-angle angle"
  :precision binary64
  (* 180.0 (/ (atan (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))) PI)))